Hill sphere

The Hill radius or sphere (the latter defined by the former radius) has been described as "the region around a planetary body where its own gravity (compared to that of the Sun or other nearby bodies) is the dominant force in attracting satellites," both natural and artificial. As described by de Pater and Lissauer, all bodies within a system such as the Sun's Solar System "feel the gravitational force of one another", and while the motions of just two gravitationally interacting bodies—constituting a "two-body problem"—are "completely integrable ([meaning]...there exists one independent integral or constraint per degree of freedom)" and thus an exact, analytic solution, the interactions of three (or more) such bodies "cannot be deduced analytically", requiring instead solutions by numerical integration, when possible. This is the case, unless the negligible mass of one of the three bodies allows approximation of the system as a two-body problem, known formally as a "restricted three-body problem". :R_{\mathrm{H}} \approx a \sqrt[3]{\frac{m_2}{3(m_1+m_2)}} \approx a \sqrt[3]{\frac{1}{3(\frac{m_1}{m_2}+1)}}, where, in this representation, semi-major axis "a" can be understood as the "instantaneous heliocentric distance" between the two masses (elsewhere abbreviated rp). :R_{\mathrm{H}} = a {\frac{(1-e^2)}{(1+e \cos(\theta))}} \sqrt[3]{\frac{m_2}{3(m_1+m_2)}} = a {\frac{(1-e^2)}{(1+e \cos(\theta))}}\sqrt[3]{\frac{1}{3(\frac{m_1}{m_2}+1)}} where \theta is the true anomaly for the less massive body in its orbit around the more massive body, an angle contained between 0° and 360°, with \cos(\theta) varying between +1 at the periapsis, \theta = 0, and -1 at the apoapsis, \theta = 180. :With (1-e^2) = (1+e)(1-e), :R_{\mathrm{H}} = a (1-e) \sqrt[3]{\frac{m_2}{3(m_1+m_2)}} = a (1-e)\sqrt[3]{\frac{1}{3(\frac{m_1}{m_2}+1)}} at the perigee and :R_{\mathrm{H}} = a (1+e) \sqrt[3]{\frac{m_2}{3(m_1+m_2)}} = a (1+e)\sqrt[3]{\frac{1}{3(\frac{m_1}{m_2}+1)}} at the apogee, in the case of the Earth-Moon system, for example, {\frac{m_2}{m_1}} = 0.0123000371(4) and {\frac{m_1}{m_2}} \approx 81.300568, a = 384,399 \ km, e = 0.05490 and the Hill radius of the Moon at the perigee is R_{\mathrm{H}} = 57,910 \ km , and at the apogee is 64,638 \ km. When eccentricity is negligible (the most favourable case for orbital stability), this expression reduces to the one presented above. == Example and derivation ==

s of each body of the Sun-Earth-Moon system. The actual Hill radius for the Moon is on the order of 60,000 km (i.e., extending less than one-sixth the distance of the 378,000 km between the Moon and the Earth). In the Earth-Sun example, the Earth () orbits the Sun () at a distance of 149.6 million km, or one astronomical unit (AU). The Hill sphere for Earth thus extends out to about 1.5 million km (0.01 AU). The Moon's orbit, at a distance of 0.384 million km from Earth, is comfortably within the gravitational sphere of influence of Earth and it is therefore not at risk of being pulled into an independent orbit around the Sun. The earlier eccentricity-ignoring formula can be re-stated as follows: :\frac{R^3_{\mathrm{H}}}{a^3} \approx 1/3 \frac{m_2}{M}, or 3\frac{R^3_{\mathrm{H}}}{a^3} \approx \frac{m_2}{M}, where M is the sum of the interacting masses. Derivation The expression for the Hill radius can be found by equating gravitational and centrifugal forces acting on a test particle (of mass much smaller than m) orbiting the secondary body. Assume that the distance between masses M and m is r, and that the test particle is orbiting at a distance r_{\mathrm{H}} from the secondary. When the test particle is on the line connecting the primary and the secondary body, the force balance requires that :\frac{Gm}{r^2_{\mathrm{H}}}-\frac{GM}{(r-r_{\mathrm{H}})^2}+\Omega^2(r-r_{\mathrm{H}})=0, where G is the gravitational constant and \Omega=\sqrt{\frac{GM}{r^3}} is the (Keplerian) angular velocity of the secondary about the primary (assuming that m\ll M). The above equation can also be written as :\frac{m}{r^2_{\mathrm{H}}}-\frac{M}{r^2}\left(1-\frac{r_{\mathrm{H}}}{r}\right)^{-2}+\frac{M}{r^2}\left(1-\frac{r_{\mathrm{H}}}{r}\right)=0, which, through a binomial expansion to leading order in r_{\mathrm{H}}/r, can be written as :\frac{m}{r^2_{\mathrm{H}}}-\frac{M}{r^2}\left(1+2\frac{r_{\mathrm{H}}}{r}\right)+\frac{M}{r^2}\left(1-\frac{r_{\mathrm{H}}}{r}\right) = \frac{m}{r^2_{\mathrm{H}}}-\frac{M}{r^2}\left(3\frac{r_{\mathrm{H}}}{r}\right)\approx 0. Hence, the relation stated above :\frac{r_{\mathrm{H}}}{r}\approx \sqrt[3]{\frac{m}{3 M}}. If the orbit of the secondary about the primary is elliptical, the Hill radius is maximum at the apocenter, where r is largest, and minimum at the pericenter of the orbit. Therefore, for purposes of stability of test particles (for example, of small satellites), the Hill radius at the pericenter distance needs to be considered. To leading order in r_{\mathrm{H}}/r, the Hill radius above also represents the distance of the Lagrangian point L1 from the secondary. == Regions of stability ==

The Hill sphere is only an approximation, and other forces such as radiation pressure or the Yarkovsky effect can eventually perturb an object out of the sphere. As stated, the satellite (third mass) should be small enough that its gravity contributes negligibly. In a two-planet system, the mutual Hill radius of the two planets must exceed 2\sqrt{3} to be stable. For systems with three or more planets, configurations in which neighbouring planets are separated by fewer than ten mutual Hill radii are inherently unstable, mainly because a third planet introduces perturbations that drain angular momentum. == Further examples ==

It is possible for a Hill sphere to be so small that it is impossible to maintain an orbit around a body. For example, an astronaut could not have orbited the 104 ton Space Shuttle at an orbit 300 km above the Earth, because a 104-ton object at that altitude has a Hill sphere of only 120 cm in radius, much smaller than a Space Shuttle. A sphere of this size and mass would be denser than lead, and indeed, in low Earth orbit, a spherical body must be more dense than lead in order to fit inside its own Hill sphere, or else it will be incapable of supporting an orbit. Satellites further out in geostationary orbit, however, would only need to be more than 6% of the density of water to fit inside their own Hill sphere. Within the Solar System, the planet with the largest Hill radius is Neptune, with 116 million km, or 0.775 au; its great distance from the Sun amply compensates for its small mass relative to Jupiter (whose own Hill radius measures 53 million km). If Planet Nine exists, however, then assuming a mass of ~10 Earths, radius of ~15,000 km, distance of ~500 AU and eccentricity of ~0.25, it would have a Hill radius of 1.2 billion km, over 10 times Neptune's Hill radius. An asteroid from the asteroid belt will have a Hill sphere that can reach 220,000 km (for 1 Ceres), diminishing rapidly with decreasing mass. The Hill sphere of 66391 Moshup, a Mercury-crossing asteroid that has a moon (named Squannit), measures 22 km in radius. A typical extrasolar "hot Jupiter", HD 209458 b, has a Hill sphere radius of 593,000 km, about eight times its physical radius of approx 71,000 km. Even the smallest close-in extrasolar planet, CoRoT-7b, still has a Hill sphere radius (61,000 km), six times its physical radius (approx 10,000 km). Therefore, these planets could have small moons close in, although not within their respective Roche limits. == Hill spheres for the Solar System ==

The following table and logarithmic plot show the radius of the Hill spheres of some bodies of the Solar System calculated with the first formula stated above (including orbital eccentricity), using values obtained from the JPL DE405 ephemeris and from the NASA Solar System Exploration website. == See also ==